C2006/F2402 '11 -- Answers to Recitation Problems #10

Having trouble with problem 1? Try the hints!

HINTS for PROB. #1
    A. In this case, you are at equil. pot. for Cl-. What does that imply? What forces are balanced? See also note.

    Note: You can look at the membrane potential as the voltage difference (potential) you will get if you set up a certain concentration difference, or you can look at it from the other end -- what if you start with a fixed voltage? Then you can consider what effect the voltage will have on the concentrations -- if you allow ions to flow. How is the voltage pushing the ions around -- is there a push in or out? In either words, given a particular voltage, due to a separation of ions, what will happen if you allow ions to flow? In this case, which way would the voltage (alone) push Cl-?

ANSWERS
1.   A. [Cl-] is higher outside. The reasoning goes like this:  The equilibrium potential refers to that membrane potential that will exactly balance the chemical force due to the concentration gradient.  The equilibrium potential for Cl- is given as -70mV, which is equal to the resting membrane potential.  So at rest, Cl- is at its equilibrium potential, or, in other words, the electrical force on Cl- is balanced by the concentration gradient.  Since the electrical force would be expected to push Cl- out of the cell, given that the outside is positive relative to the inside, the electrical force must be balanced by a chemical force pushing Cl- into the cell, implying that the concentration is higher outside the cell.

   If you plug into the Nernst Equation you get:     -70 = -58 log [Cl-]_out/[Cl-]_in. 

Notice the minus sign on the right because Cl- is a negatively charged ion (Z = charge on ion = -1, not +1).

Now you can solve for the ratio of [Cl-]_out to [Cl-]_in:         1.2 = log [Cl-]_out/[Cl-]_in

[Cl-]_out/[Cl-]_in = 15.8 = about 16X as much Cl- out as in.

    B-1.  Cl- ions. Hyperpolarization (towards more negative values)can be caused by ion flow from opening of either Cl- or K⁺ channels.  In this case, it must be Cl- channels, because there is no change in polarization when resting potential is same as equilibrium potential for Cl-. When chemical and electrical gradients for Cl- are balanced (= equilibrium potential), then there is no net flow of Cl- when Cl- channels are opened. (If it were opening of K⁺ channels that hyperpolarized the cell, the results would be different. When you opened the K⁺ channels at an RP of -70 mV, K⁺ would not be at equilibrium, and K⁺ would flow out of the cell, toward its equilibrium. However, at -90 mV, K⁺ would be at equilibrium, and no change would occur when you opened the channels.)
   
    B-2. Negative ions going out. We know from part B-1 that it is chloride channels that are opening here. At -70 mV, the electrical and chemical gradients are balanced, and there is no net movement of ions.  Above the equilibrium value, the cell becomes more negative in response to transmitter, so Cl- must move into the cell. In other words, the chemical gradient must prevail and push chloride in, hyperpolarizing the cell. Below the equilibrium value, the electrical gradient must prevail and push chloride out, depolarizing the cell. (If  K⁺ channels were opening here, then K⁺ would have to go in the opposite direction -- out to hyperpolarize and in to depolarize.)

   B-3. Ions going up their concentration gradient. Note that the ions go in opposite directions at different starting RP values. So the direction that they move must depend on both the chemical and electrical gradients. At -100 mV, the chemical force driving ions down their concentration gradients is less than the electrical force (potential) driving ions in the opposite direction. Either Cl- is forced out, or K⁺ is forced in, against their respective concentration gradients, making the cell more positive and depolarizing it. (We know it is really the chloride that is moving,  but if you thought it was potassium that moves, then the K⁺ must go in, up its gradient.)

2. A-1. The AcCh receptor in cardiac muscle cells is (metabotropic).

    A-2. This receptor uses (no second messenger).

    A-3. Add AcCh to (the extracellular side) AND beta/gamma subunit to cyto side.

     A-4. Need detached patches to check the effects of (the G protein or its parts) .

 In this case the G protein itself (actually its beta-gamma subunit) opens the channel. No second messenger is involved.

AcCh binds to the extracellular domain of the the transmembrane receptor. (So AcCh must be added from the extracellular side, and can be added to an attached patch.) The cytoplasmic domain of the receptor changes in conformation and activates the G protein. The G protein exchanges GDP for GTP and comes apart into the alpha  subunit (bound to GTP) and the beta-gamma subunit. The beta-gamma subunit binds to the cytoplasmic domain of the channel protein and opens the channel. Since the beta-gamma subunit acts from the inside of the cell, a detached patch is needed to test its effects.

     B. You will block the current if you add (alpha subunit with GDP).

 The alpha subunit hydrolyzes the GTP to GDP, and the alpha-GDP subunit rebinds to the beta-gamma subunit. This inactivates the beta-gamma subunit so it can no longer bind to the cytoplasmic domain of the channel. Without the beta-gamma subunit, the channel closes. (When the channels are open, there is actually an equilibrium between beta-gamma subunits bound to the channels and beta-gamma subunits in the cytosol. At any one time, not all the beta-gamma subunits are bound to the channels. As soon as the alpha subunit hydrolyzes its GTP to GDP, the alpha-GDP binds to the free beta-gamma subunits, shifting the equilibrium away from beta-gamma subunits bound to the channels. As the free beta-gamma subunits bind to alpha-GDP subunits, more beta-gamma subunits separate from the channels, and the channels close.)